Integrand size = 25, antiderivative size = 73 \[ \int \frac {1}{\left (3-x+2 x^2\right ) \left (2+3 x+5 x^2\right )} \, dx=\frac {3 \arctan \left (\frac {1-4 x}{\sqrt {23}}\right )}{22 \sqrt {23}}+\frac {13 \arctan \left (\frac {3+10 x}{\sqrt {31}}\right )}{22 \sqrt {31}}-\frac {1}{44} \log \left (3-x+2 x^2\right )+\frac {1}{44} \log \left (2+3 x+5 x^2\right ) \] Output:
3/506*arctan(1/23*(1-4*x)*23^(1/2))*23^(1/2)+13/682*arctan(1/31*(3+10*x)*3 1^(1/2))*31^(1/2)-1/44*ln(2*x^2-x+3)+1/44*ln(5*x^2+3*x+2)
Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (3-x+2 x^2\right ) \left (2+3 x+5 x^2\right )} \, dx=-\frac {3 \arctan \left (\frac {-1+4 x}{\sqrt {23}}\right )}{22 \sqrt {23}}+\frac {13 \arctan \left (\frac {3+10 x}{\sqrt {31}}\right )}{22 \sqrt {31}}-\frac {1}{44} \log \left (3-x+2 x^2\right )+\frac {1}{44} \log \left (2+3 x+5 x^2\right ) \] Input:
Integrate[1/((3 - x + 2*x^2)*(2 + 3*x + 5*x^2)),x]
Output:
(-3*ArcTan[(-1 + 4*x)/Sqrt[23]])/(22*Sqrt[23]) + (13*ArcTan[(3 + 10*x)/Sqr t[31]])/(22*Sqrt[31]) - Log[3 - x + 2*x^2]/44 + Log[2 + 3*x + 5*x^2]/44
Time = 0.25 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {1311, 27, 1142, 25, 1083, 217, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\left (2 x^2-x+3\right ) \left (5 x^2+3 x+2\right )} \, dx\) |
\(\Big \downarrow \) 1311 |
\(\displaystyle \frac {1}{242} \int -\frac {11 (2 x+1)}{2 x^2-x+3}dx+\frac {1}{242} \int \frac {11 (5 x+8)}{5 x^2+3 x+2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{22} \int \frac {5 x+8}{5 x^2+3 x+2}dx-\frac {1}{22} \int \frac {2 x+1}{2 x^2-x+3}dx\) |
\(\Big \downarrow \) 1142 |
\(\displaystyle \frac {1}{22} \left (-\frac {3}{2} \int \frac {1}{2 x^2-x+3}dx-\frac {1}{2} \int -\frac {1-4 x}{2 x^2-x+3}dx\right )+\frac {1}{22} \left (\frac {13}{2} \int \frac {1}{5 x^2+3 x+2}dx+\frac {1}{2} \int \frac {10 x+3}{5 x^2+3 x+2}dx\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{22} \left (\frac {1}{2} \int \frac {1-4 x}{2 x^2-x+3}dx-\frac {3}{2} \int \frac {1}{2 x^2-x+3}dx\right )+\frac {1}{22} \left (\frac {13}{2} \int \frac {1}{5 x^2+3 x+2}dx+\frac {1}{2} \int \frac {10 x+3}{5 x^2+3 x+2}dx\right )\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {1}{22} \left (\frac {1}{2} \int \frac {1-4 x}{2 x^2-x+3}dx+3 \int \frac {1}{-(4 x-1)^2-23}d(4 x-1)\right )+\frac {1}{22} \left (\frac {1}{2} \int \frac {10 x+3}{5 x^2+3 x+2}dx-13 \int \frac {1}{-(10 x+3)^2-31}d(10 x+3)\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {1}{22} \left (\frac {1}{2} \int \frac {1-4 x}{2 x^2-x+3}dx-\frac {3 \arctan \left (\frac {4 x-1}{\sqrt {23}}\right )}{\sqrt {23}}\right )+\frac {1}{22} \left (\frac {1}{2} \int \frac {10 x+3}{5 x^2+3 x+2}dx+\frac {13 \arctan \left (\frac {10 x+3}{\sqrt {31}}\right )}{\sqrt {31}}\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle \frac {1}{22} \left (-\frac {3 \arctan \left (\frac {4 x-1}{\sqrt {23}}\right )}{\sqrt {23}}-\frac {1}{2} \log \left (2 x^2-x+3\right )\right )+\frac {1}{22} \left (\frac {13 \arctan \left (\frac {10 x+3}{\sqrt {31}}\right )}{\sqrt {31}}+\frac {1}{2} \log \left (5 x^2+3 x+2\right )\right )\) |
Input:
Int[1/((3 - x + 2*x^2)*(2 + 3*x + 5*x^2)),x]
Output:
((-3*ArcTan[(-1 + 4*x)/Sqrt[23]])/Sqrt[23] - Log[3 - x + 2*x^2]/2)/22 + (( 13*ArcTan[(3 + 10*x)/Sqrt[31]])/Sqrt[31] + Log[2 + 3*x + 5*x^2]/2)/22
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Int[1/(((a_) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_) + (e_.)*(x_) + (f_.)*(x_)^2 )), x_Symbol] :> With[{q = c^2*d^2 - b*c*d*e + a*c*e^2 + b^2*d*f - 2*a*c*d* f - a*b*e*f + a^2*f^2}, Simp[1/q Int[(c^2*d - b*c*e + b^2*f - a*c*f - (c^ 2*e - b*c*f)*x)/(a + b*x + c*x^2), x], x] + Simp[1/q Int[(c*e^2 - c*d*f - b*e*f + a*f^2 + (c*e*f - b*f^2)*x)/(d + e*x + f*x^2), x], x] /; NeQ[q, 0]] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[e^2 - 4*d*f, 0]
Time = 3.07 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.82
method | result | size |
default | \(-\frac {\ln \left (2 x^{2}-x +3\right )}{44}-\frac {3 \sqrt {23}\, \arctan \left (\frac {\left (4 x -1\right ) \sqrt {23}}{23}\right )}{506}+\frac {\ln \left (5 x^{2}+3 x +2\right )}{44}+\frac {13 \arctan \left (\frac {\left (10 x +3\right ) \sqrt {31}}{31}\right ) \sqrt {31}}{682}\) | \(60\) |
risch | \(-\frac {3 \sqrt {23}\, \arctan \left (\frac {\left (4 x -1\right ) \sqrt {23}}{23}\right )}{506}-\frac {\ln \left (16 x^{2}-8 x +24\right )}{44}+\frac {\ln \left (100 x^{2}+60 x +40\right )}{44}+\frac {13 \arctan \left (\frac {\left (10 x +3\right ) \sqrt {31}}{31}\right ) \sqrt {31}}{682}\) | \(60\) |
Input:
int(1/(2*x^2-x+3)/(5*x^2+3*x+2),x,method=_RETURNVERBOSE)
Output:
-1/44*ln(2*x^2-x+3)-3/506*23^(1/2)*arctan(1/23*(4*x-1)*23^(1/2))+1/44*ln(5 *x^2+3*x+2)+13/682*arctan(1/31*(10*x+3)*31^(1/2))*31^(1/2)
Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\left (3-x+2 x^2\right ) \left (2+3 x+5 x^2\right )} \, dx=\frac {13}{682} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) - \frac {3}{506} \, \sqrt {23} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) + \frac {1}{44} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) - \frac {1}{44} \, \log \left (2 \, x^{2} - x + 3\right ) \] Input:
integrate(1/(2*x^2-x+3)/(5*x^2+3*x+2),x, algorithm="fricas")
Output:
13/682*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) - 3/506*sqrt(23)*arctan(1 /23*sqrt(23)*(4*x - 1)) + 1/44*log(5*x^2 + 3*x + 2) - 1/44*log(2*x^2 - x + 3)
Time = 0.12 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.14 \[ \int \frac {1}{\left (3-x+2 x^2\right ) \left (2+3 x+5 x^2\right )} \, dx=- \frac {\log {\left (x^{2} - \frac {x}{2} + \frac {3}{2} \right )}}{44} + \frac {\log {\left (x^{2} + \frac {3 x}{5} + \frac {2}{5} \right )}}{44} - \frac {3 \sqrt {23} \operatorname {atan}{\left (\frac {4 \sqrt {23} x}{23} - \frac {\sqrt {23}}{23} \right )}}{506} + \frac {13 \sqrt {31} \operatorname {atan}{\left (\frac {10 \sqrt {31} x}{31} + \frac {3 \sqrt {31}}{31} \right )}}{682} \] Input:
integrate(1/(2*x**2-x+3)/(5*x**2+3*x+2),x)
Output:
-log(x**2 - x/2 + 3/2)/44 + log(x**2 + 3*x/5 + 2/5)/44 - 3*sqrt(23)*atan(4 *sqrt(23)*x/23 - sqrt(23)/23)/506 + 13*sqrt(31)*atan(10*sqrt(31)*x/31 + 3* sqrt(31)/31)/682
Time = 0.11 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\left (3-x+2 x^2\right ) \left (2+3 x+5 x^2\right )} \, dx=\frac {13}{682} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) - \frac {3}{506} \, \sqrt {23} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) + \frac {1}{44} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) - \frac {1}{44} \, \log \left (2 \, x^{2} - x + 3\right ) \] Input:
integrate(1/(2*x^2-x+3)/(5*x^2+3*x+2),x, algorithm="maxima")
Output:
13/682*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) - 3/506*sqrt(23)*arctan(1 /23*sqrt(23)*(4*x - 1)) + 1/44*log(5*x^2 + 3*x + 2) - 1/44*log(2*x^2 - x + 3)
Time = 0.11 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.81 \[ \int \frac {1}{\left (3-x+2 x^2\right ) \left (2+3 x+5 x^2\right )} \, dx=\frac {13}{682} \, \sqrt {31} \arctan \left (\frac {1}{31} \, \sqrt {31} {\left (10 \, x + 3\right )}\right ) - \frac {3}{506} \, \sqrt {23} \arctan \left (\frac {1}{23} \, \sqrt {23} {\left (4 \, x - 1\right )}\right ) + \frac {1}{44} \, \log \left (5 \, x^{2} + 3 \, x + 2\right ) - \frac {1}{44} \, \log \left (2 \, x^{2} - x + 3\right ) \] Input:
integrate(1/(2*x^2-x+3)/(5*x^2+3*x+2),x, algorithm="giac")
Output:
13/682*sqrt(31)*arctan(1/31*sqrt(31)*(10*x + 3)) - 3/506*sqrt(23)*arctan(1 /23*sqrt(23)*(4*x - 1)) + 1/44*log(5*x^2 + 3*x + 2) - 1/44*log(2*x^2 - x + 3)
Time = 16.72 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.08 \[ \int \frac {1}{\left (3-x+2 x^2\right ) \left (2+3 x+5 x^2\right )} \, dx=\ln \left (x-\frac {1}{4}-\frac {\sqrt {23}\,1{}\mathrm {i}}{4}\right )\,\left (-\frac {1}{44}+\frac {\sqrt {23}\,3{}\mathrm {i}}{1012}\right )-\ln \left (x-\frac {1}{4}+\frac {\sqrt {23}\,1{}\mathrm {i}}{4}\right )\,\left (\frac {1}{44}+\frac {\sqrt {23}\,3{}\mathrm {i}}{1012}\right )-\ln \left (x+\frac {3}{10}-\frac {\sqrt {31}\,1{}\mathrm {i}}{10}\right )\,\left (-\frac {1}{44}+\frac {\sqrt {31}\,13{}\mathrm {i}}{1364}\right )+\ln \left (x+\frac {3}{10}+\frac {\sqrt {31}\,1{}\mathrm {i}}{10}\right )\,\left (\frac {1}{44}+\frac {\sqrt {31}\,13{}\mathrm {i}}{1364}\right ) \] Input:
int(1/((2*x^2 - x + 3)*(3*x + 5*x^2 + 2)),x)
Output:
log(x - (23^(1/2)*1i)/4 - 1/4)*((23^(1/2)*3i)/1012 - 1/44) - log(x + (23^( 1/2)*1i)/4 - 1/4)*((23^(1/2)*3i)/1012 + 1/44) - log(x - (31^(1/2)*1i)/10 + 3/10)*((31^(1/2)*13i)/1364 - 1/44) + log(x + (31^(1/2)*1i)/10 + 3/10)*((3 1^(1/2)*13i)/1364 + 1/44)
Time = 0.18 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\left (3-x+2 x^2\right ) \left (2+3 x+5 x^2\right )} \, dx=\frac {13 \sqrt {31}\, \mathit {atan} \left (\frac {10 x +3}{\sqrt {31}}\right )}{682}-\frac {3 \sqrt {23}\, \mathit {atan} \left (\frac {4 x -1}{\sqrt {23}}\right )}{506}+\frac {\mathrm {log}\left (5 x^{2}+3 x +2\right )}{44}-\frac {\mathrm {log}\left (2 x^{2}-x +3\right )}{44} \] Input:
int(1/(2*x^2-x+3)/(5*x^2+3*x+2),x)
Output:
(598*sqrt(31)*atan((10*x + 3)/sqrt(31)) - 186*sqrt(23)*atan((4*x - 1)/sqrt (23)) + 713*log(5*x**2 + 3*x + 2) - 713*log(2*x**2 - x + 3))/31372